Circle $C$ with radius 2 has diameter $\overline{AB}$. Circle D is internally tangent to circle $C$ at $A$. Circle $E$ is internally tangent to circle $C$, externally tangent to circle $D$, and tangent to $\overline{AB}$. The radius of circle $D$ is three times the radius of circle $E$, and can be written in the form $\sqrt{m}-n$, where $m$ and $n$ are positive integers. Find $m+n$.

Answer: [asy] import graph; size(7.99cm);  real labelscalefactor = 0.5;  pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black;  real xmin = 4.087153740193288, xmax = 11.08175859031552, ymin = -4.938019122704778, ymax = 1.194137062512079;  draw(circle((7.780000000000009,-1.320000000000002), 2.000000000000000));  draw(circle((7.271934046987930,-1.319179731427737), 1.491933384829670));  draw(circle((9.198812158392690,-0.8249788848962227), 0.4973111282761416));  draw((5.780002606580324,-1.316771019595571)--(9.779997393419690,-1.323228980404432));  draw((9.198812158392690,-0.8249788848962227)--(9.198009254448635,-1.322289365031666));  draw((7.271934046987930,-1.319179731427737)--(9.198812158392690,-0.8249788848962227));  draw((9.198812158392690,-0.8249788848962227)--(7.780000000000009,-1.320000000000002));  dot((7.780000000000009,-1.320000000000002),dotstyle);  label("$C$", (7.707377218471464,-1.576266740352400), NE * labelscalefactor);  dot((7.271934046987930,-1.319179731427737),dotstyle);  label("$D$", (7.303064016111554,-1.276266740352400), NE * labelscalefactor);  dot((9.198812158392690,-0.8249788848962227),dotstyle);  label("$E$", (9.225301294671791,-0.7792624249832147), NE * labelscalefactor);  dot((9.198009254448635,-1.322289365031666),dotstyle);  label("$F$", (9.225301294671791,-1.276266740352400), NE * labelscalefactor);  dot((9.779997393419690,-1.323228980404432),dotstyle);  label("$B$", (9.810012253929656,-1.276266740352400), NE * labelscalefactor);  dot((5.780002606580324,-1.316771019595571),dotstyle);  label("$A$", (5.812051070003994,-1.276266740352400), NE * labelscalefactor);  clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]
Using the diagram above, let the radius of $D$ be $3r$, and the radius of $E$ be $r$. Then, $EF=r$, and $CE=2-r$, so the Pythagorean theorem in $\triangle CEF$ gives $CF=\sqrt{4-4r}$. Also, $CD=CA-AD=2-3r$, so\[DF=DC+CF=2-3r+\sqrt{4-4r}.\]Noting that $DE=4r$, we can now use the Pythagorean theorem in $\triangle DEF$ to get\[(2-3r+\sqrt{4-4r})^2+r^2=16r^2.\]
Solving this quadratic is somewhat tedious, but the constant terms cancel, so the computation isn't terrible. Solving gives $3r=\sqrt{240}-14$ for a final answer of $\boxed{254}$.
Notice that C, E and the point of tangency to circle C for circle E will be concurrent because C and E intersect the tangent line at a right angle, implying they must be on the same line.